Strongly collectionwise normal spaces with $G_\delta$ diagonal are submetrizable

Indeed, any strongly collectionwise normal space with $G_\delta$ diagonal is submetrizable. To see this, lets follow the proof in Borges's paper and check that it's still valid with those weaker assumptions. Borges also uses references instead of showing everything explicitly, so it will also be helpful to write it out.

Here $R^n$ for $R\subseteq X\times X$ is recursively defined by $R^1 = R$ and $R^{n+1} = R\circ R^n$ where $\circ$ denotes composition of relations.

First let $V_n\subseteq X\times X$ be open such that $\Delta_X = \bigcap_n V_n$. From strong collectionwise normality, pick open symmetric neighbourhoods of the diagonal $U_n$ such that $U_{n+1}^2\subseteq U_n$ and $U_n\subseteq V_n$ (see proposition 3.11 in Normal topological spaces by Alo and Shapiro for why it's possible).

Note that because $U_n$ are symmetric, contain the diagonal, and $U_{n+1}^2\subseteq U_n$, the set $\{U_n : n\in\mathbb{N}\}$ is a base for a uniformity $\mathcal{U}$ on $X$ (with $X$ treated as a set). Denote this uniform space as $M$. Since $\Delta_X = \bigcap_n V_n$ and $\Delta_X\subseteq U_n\subseteq V_n$, it follows that $\bigcap\mathcal{U} = \Delta_X$ so that $M$ is Hausdorff.

Since $M$ is a uniform space with countable basis for its uniformity, it's pseudometrizable, and so metrizable. This is a well-know theorem, and I'll show how it can be seen using the following theorem 8.2 from Normal topological spaces:

Theorem. Suppose that $(\mathcal{A}_n)$ is a sequence of open covers of $Y$ such that $\mathcal{A}_{n+1}$ is a $2$-refinement of $\mathcal{A}_n$, then there exists a continuous pseudometric $d$ on $Y$ such that $\text{St}(x, \mathcal{A}_{n+2})\subseteq B_d(x, 1/2^n)\subseteq \text{St}(x, \mathcal{A}_n)$.

Here $2$-refinement means that if $U, V\in\mathcal{A}_{n+1}$ and $U\cap V\neq\emptyset$ then $U\cap V\subseteq W$ for some $W\in\mathcal{A}_n$.

For any element $U\in\mathcal{U}$ one can take open symmetric $W_n\in\mathcal{U}$ such that $W_{n+1}^2\subseteq W_n$ and $W_1\subseteq U$. This is because if $V\in\mathcal{U}$ then $\text{int}(V)\in\mathcal{U}$. Then the sets $W_n(y) = \{x\in M : (x, y)\in W_n\}$ are such that $\mathcal{A}_n = \{W_n(y) : y\in M\}$ are open covers of $M$, and if $W_{n+1}(y)\cap W_{n+1}(z)\neq \emptyset$ then there is $x\in M$ with $(x, y), (x, z)\in W_{n+1}$. If $w\in W_{n+1}(y)$ for example, then $(w, y)\in W_{n+1}$ so $(w, x)\in W_n$ and so $w\in W_n(x)$. This shows that $W_{n+1}(y)\cup W_{n+1}(z)\subseteq W_n(x)$ and so $\mathcal{A}_{n+1}$ $2$-refines $\mathcal{A}_n$. So there is a continuous pseudometric $d$ on $M$ with $B_d(x, 1)\subseteq \text{St}(x, W_2)\subseteq W_1(x)\subseteq U(x)$ for all $x\in M$.

If we find continuous pseudometrics $d_n$ with $d_n\leq 1$ and $B_{d_n}(x, 1)\subseteq U_n(x)$ for all $x\in M$, let $d = \sum_n 2^{-n}d_n$. Then $d$ is a continuous pseudometric on $M$. If $O\subseteq M$ is open and $x\in O$, then $U_n(x)\subseteq O$ for some $n$, and so $B_d(x, 2^{-n})\subseteq B_{d_n}(x, 1)\subseteq U_n(x)\subseteq O$, and so $O$ is $d$-open. And since $d$ is continuous, any $d$-open set is open in $M$. So $M$ is pseudometrizable.

Now all is left to show is that $i:X\to M$ given by $i(x) = x$ is a continuous map. Suppose that $x\in O$ where $O\subseteq M$ is open. Then $x\in U_n(x)\subseteq O$ and $U_n(x)$ is open in $X$. This shows $i$ is continuous, and so $X$ is submetrizable.